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asambeis [7]
3 years ago
10

To determine the trajectory of the movement of a material point in the xOz plane that is subjected simultaneously to oscillation

s x (t) = A * sin π(t + 1) and y (t) = A * sin (πt + π / 2).
Physics
1 answer:
nasty-shy [4]3 years ago
4 0

We are to show that the given parametric curve is a circle.

The trajectory of a circle with a radius r will satisfy the following relationship:

(x-x_c)^2 + (y-y_c)^2 = r^2

(with (x_c,y_c) being the center point)

We are given the x and y in a parametric form which can be further rewritten (using properties of sin/cos):

x(t) = A\sin \pi(t+1) = A\sin (\pi t + \pi) = -A\sin \pi t\\y(t) = A\sin (\pi t + \frac{\pi}{2}) = A\cos \pi t

Squaring and adding both gives:

x^2(t) + y^2(t) = A^2(-1)^2\sin^2 \pi t + A^2 \cos^2 \pi t = A^2\\\implies x^2 + y^2 = A^2

The last expression shows that the given parametric curve is a circle with the center (0,0) and radius A.


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If the inside of a container has a surface area of 20cm², what will be the pressure on each square inch of the container if 117.
egoroff_w [7]

Answer:

Explanation:

given :-

force applied = 117.6N

Area = 20 m^2

Pressure =?

solution:-

Pressure = Force / Area

= 117.6 N / 20 m^2

= 5.88 Pascals

Hope it helps :D!

Explanation:

6 0
2 years ago
A weather reporter describes a large area of air that has the same temperature and air pressure . What is the reporter describin
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<u>Answer:</u>

<u>air mass</u>

<u>Explanation:</u>

Remember, the weather of a particular location is often a description of the atmospheric condition of that particular place.

Indeed, the definition of an air mass fits the reporter's description above. As the name implies, <u>a particular air mass would represent the temperature characteristics of a particular area that has a constant or the same temperature and air pressure.</u>

5 0
3 years ago
two soccer players, mary and jane, begin running from nearly the same point at the same time. mary runs in an easterly direction
goblinko [34]

The time taken before the two soccer players are 24.7 m apart is 3 seconds.

Velocity of Jane relative to marry is 4.4 m/s in y-direction and 6.95 m/s in x direction.

The distance between Jane and Mary after 4.08 seconds is 33.58 m.

<h3>Resultant velocity of Mary and Jane</h3>

Vertical component of the velocity;

Jane: Vy1 = V sinθ

Vy1 = 5.07 m/s  x  sin(60.2)

Vy1 = 4.4 m/s

Horizontal components of the velocity;

Jane: Vx1 = V cos θ

Vx1 = 5.07 m/s  x  cos60.2

Vx1 = 2.52 m/s

mary: Vx2 = 4.43 m/s

total horizontal components of the velocity = 4.43 m/s + 2.52 m/s = 6.95 m/s

<h3>Resultant velocity</h3>

V = √Vy² + Vx²

V = √(4.4² + 6.95²)

V = 8.23 m/s

<h3>Time taken before they are 24.7 m apart</h3>

t = 24.7 m / 8.23 m/s

t = 3 seconds

Velocity of Jane relative to marry is 4.4 m/s in y-direction and 6.95 m/s in x direction.

<h3>Distance between Jane and Mary after 4.08 seconds</h3>

d = vt

d = 8.23 m/s  x 4.08 s

d = 33.58 m

Learn more about relative velocity here: brainly.com/question/17228388

#SPJ1

5 0
1 year ago
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